Isomorphism proof

3. The attempt at a solution
To start,
I wrote out all of the elements of [tex]Z_2\oplus Z_4[/tex]. There are 8 of them, of course. Then I need to find the automorphisms of it. It looks to me like they would be the same as Aut(D8),which I have. I have proved (I think) that [tex]Aut(D_8)\cong D_8[/tex]. I'm just unsure HOW to go about this one.

Could I show with a cayley table that [tex]Z_2\oplus Z_4[/tex] is nonabelian and it's clearly of order 8, so then the isomorphism holds?

I just get stuck on these types of problems. the idea of isomorphisms is still pretty fresh to me and so are the groups themselves.

Just because the orders match up doesn't mean that the groups are isomorphic. D8 is non-abelian, Z2 x Z4 is abelian. They are NOT isomorphic. But anyways, if X is isomorphic to Y, and Aut(X) is isomorphic to Z, then Aut(Y) is isomorphic to Z.

Well, if you have no idea where to start, then you might try listing all the automorphisms, and then worry about figuring out what group they form. To do this, note that to specifiy an isomorphism (in fact, any homomorphism), it suffices to specify it on a generating set. In this case, we could take, eg, (1,0) and (0,1). These must be mapped to another pair of generators of order 2 and 4 respectively, and it can be shown this condition is also sufficient, so it remains to find all such pairs. Since there are 3 elements of order 2 and 4 of order 4, there are apparently 8 pairs that generate the group and 4 that don't. For example, (0,2),(0,1) doesn't.

OK,
I don't mean to be dense here, but I'm just not GETTING this.
It makes sense to me that (1,0),(0,1) generates the group because 1 generates Z2 and 1 also generates Z4, so putting that together, you get [tex]Z_2\oplus Z_4[/tex]
My confusion comes when I try to find the other generating pairs.

I think that (1,0),(0,3) is a generating pair by the same logic as above.

(0,1),(1,2) is confusing me. I see that 1 generates Z2, then I have for Z4 both 1 and 2. Now 1 generates Z4, but 2 doesn't. Does the overlap matter?

Other pairs I came up with:
(0,3)(1,2)----->generates Z2 once,Z4 once, subgroup of Z4 as well
(1,1)(1,0)----->generates Z2 twice, Z4 once
(1,1)(1,2)----->generates Z2 twice,Z4 once, subgroup of Z4 as well
(1,1)(0,2)----->generates Z2 once, Z4 once, subgroup of Z4 as well
(1,3)(1,0)----->generates Z2 twice, Z4 once
(1,3)(1,2)----->generates Z2 twice, Z4 once, subgroup of Z4 as well
(1,3)(0,2)----->generates Z2 once, Z4 once, subgroup of Z4 as well

In this case, you can just check by writing out the options.
For example, (0, 1) and (1, 2) generate the group, because by adding (0, 1)'s I can make {0} x Z4.
I can make Z2 x {0}, it is {(1, 2)+2(0,1), 2(1, 2)}
In fact, any element (n, m) I can create by first taking 1 or 2 times (1, 2) to get the first component right, and then adding (0, 1) as often as I want to get the second correct (it doesn't change the first one anymore).

Try to understand this. If you think you have it, try to see why this works (and also (1, 2),(0,3) work, and (1,1),(0,3) does as well, but (1, 2), (1, 0) doesn't).

The trick is to think about the entire thing at once as pairs of numbers, which I can just add and then take modulo 2 in the first, modulo 4 in the second component; rather than as separate groups which I have to generate first and put together then.

OK
is this correct?
11, 02=(1,3)
01,10=(1,1)
03,10=(1,3)
03,12=(1,1)
13,02=(1,1)
01,12=(1,1)
and each one of those are generators for the group (I THINK) I get an ordered pair that generates both sets.
these are the ones I'm not sure of:
11,12=(0,3)....+(1,1)=(0,0) but I don't think I can DO that here..I'm changing both entries...
13,10=(0,3)...+(1,0)=(1,3)..I *think* that works?
11,10=(0,1)....+(1,0)=(1,1)..I *think* that works?
13,12=(2,5)=(0,1)+(1,2)=(1,3)...don't know about that, changing both entries.

so with the 6 at the top and the 2 i *think* work, am I getting this at all?

To see if a pair (a,b), (c,d) generates the group, note that this means precisely that for any (x,y) there exist integers n,m such that:

n(a,b)+m(c,d) = (x,y)

ie, as integers, you need simultaneous solutions to the equations:

na+mc=x (mod 2)
nb+md=y (mod 4)

The second equation tells you that at least one of b,d must be 1 or 3, since otherwise you'll only ever get even numbers. Whichever pair contains the 1 or 3 will necessarily be of order 4, so the other one must be of order 2. We can assume b is 1 or 3, so (a,b) is of order 4. Then for (c,d) to be of order 2, d must be 0 or 2.

Now just go through it. If b=1,d=0, the solutions are exactly those with n=y (mod 4), m= anything. You need to pick a and c so that among these n and m (for any y) you can obtain all the solutions x. For example, a=1,c=1 will work, but a=1,c=0 will not.

You can go from here either by brute force, or by applying more logical thinking to reduce the amount of mindless checking you need to do. In a problem like this with small numbers, brute force is usually easier, but for problems with bigger numbers, or (god forbid) letters instead of numbers, you'll have to do every step logically and abstractly, and it doesn't hurt to get used to it.

Hi,
I really appreciate your input on this problem.
I still don't know if my thought process is right or wrong.
Please tell me if my work is right or wrong.

I HAVE tried to think it through in a logical step by step manner. I've been stuck on this problem for 3 days. I'm frustated. I just don't understand this. I thought that I WAS using brute force. I thought that I FINALLY was understanding this automorphism thing...and I'm STILL wrong. I think that I've simply looked at this thing too long for it to make any sense to me.

I haven't ever had any algebraic strructures or group theory,ever.

I know it's easy to you guys. I am a more concrete thinker and I'm having a difficult time relating all of this to things that my mind likes to think about.

I give up on this for another day.
I'll get some more paper out and try this again later.

BTW,
are ANY of the pairs I came up with correct?
Thanks,

CC
....the capitol letters aren't meant to be *shouty*..I'm just SOOOO very frustrated with this problem...

Really, I'm not sure what you're thought processes are. I'm sure it will help at least one of us if you took a deep breath and started from at the very top.

Incidentally, my idea of a brute force approach would be to:

Let G = Z2 x Z4.

(1) Write down all candidates for a homomorphism G -> G.
(2) Check to see which are homomorphisms.
(3) Check to see which are automorphisms.
(4) Compute the multiplication table on the automorphisms.
(5) Compute the multiplication table for D8.
(5) Find a permutation of the elements of Aut(G) that makes the multiplication tables the same.

I suppose this isn't the only brute force approach one might take, though.

(By "candidate", I'm thinking of the theorem that any homomorphism is uniquely determined by its value on a generating set. Since we've chosen a generating set with 2 elements, a candidate for a homomorphism is simply a choice of an image for each generator)

Hello,
I am working on this some more. I have had time to carefully go through everything again and I understand HOW to find the generators now.
I got
(10)(01)
(01)(12)
(03)(10)
(11)(10)
(11)(12)
(13)(10)
(13)(12)
Now I need to permute them to make the table the same, right?
I have the table for D8, and I'm not quite sure how to proceed here. I feel like getting the generators...and understanding WHY they work was a huge step for me....Do I just start out and say(01)--->(03) and just go from there?
Help me move forward here!
Thanks
CC

(1) (1,0) and (0,1) is a generating set for Z2xZ4.
(2) Any homomorphism is uniquely determined by its value on any generating set.
(3) Automorphisms preserve the orders of elements.
(4) For finite groups, a homomorphism is surjective iff it's an automorphism.

Together, these tell you that the automorphisms of Z2xZ4 correspond uniquely to pairs of elements of Z2xZ4 that satisfy:
(a) The first element has order 2
(b) The second element has order 4
(c) Together, the two elements generate Z2xZ4

You then proceeded to find all pairs of elements of Z2xZ4 that satisfied conditions (a), (b), and (c).

I'm unclear about how you went on to show that Aut(G) is congruent to D8. First, if we let

Hi,
I had proved (with my professors' help) that Aut(D8)=D8. I have the table of all of those automorphisms, so it wasn't too difficult to write out the automorphisms for Z2 x Z4, just replacing the ordered pairs with the letters that I used in that chart.
I let (01)(10) be r a, where r is degree 4 and a is degree 2, generators for D8 corresponding to the 2 generators for Aut(Z2 x Z4). I just proceeded in the same fashion for each pair and got the permutation table that matches AUT(D8) and since I know that AUT(D8)=D8, I'm there.

I did show my work to the professor, and he agreed that it was ("the sledge hammer method") but valid.

So I guess I'm OK. At least the problem taught me a LOT. I'm glad I got it finally.